In the Stochastic Calculus of Durrett Example 3.2.
I understand the usage of strong markov property. But I am confused about how $B_T \circ \theta_S=B_T$. Since $B_T \circ \theta_S(\omega(t))=B_T(\omega(t+S))$, is $T$ also shifted here? Because otherwise these two seems not equal. However, my professor says $T$ is not shifted, so I am very confused. Could anyone help explain this? Thanks!
Since $T$ is a hitting time and $T(\omega)>S(\omega)$ we can write $T(\omega)=S(\omega)+T(\theta_S\omega)$. Expressing the Brownian motion explicitly as a function of two variables, i.e., $B_T(\omega)=B(\omega,T(\omega))$ we have $$B_T(\theta_S\omega)=B(\theta_S\omega,T(\theta_S\omega)) =B(\omega, S(\omega)+T(\theta_S\omega))=B(\omega,T(\omega))=B_T(\omega).$$